Could anyone help me derive Bohr's purely classical result for the angular velocity of an orbiting electron

w=( (E^3/2)*(m^-1/2) ) / (2^3/2)*c

w= angular velocity, E=Total energy, m=mass of electron, c= a constant (not speed of light)

Question

Could anyone help me derive Bohr's  purely classical result for the angular velocity of an orbiting electron 

w=( (E^3/2)*(m^-1/2) ) / (2^3/2)*c 

w= angular velocity, E=Total energy, m=mass of electron, c= a constant (not speed of light)

anonymous · Answer

Which is formally equivalent to Kepler’s Third Law of planetary motion.

anonymous · Answer

c=the speed of light is actually the speed of light*

can get, w=v/r - (1)

             r=c/2E, - (2)

             E=(mv^2)/2 - (3)

Rearrange (3) for v and sub (3) and (2) back in to (1) and gets

            w=( (E^3/2)*(m^-1/2) )*(2^3/2) ) / c 

As you can see, end up with 2^3/2 on top instead of on the bottom. Help?

anonymous · Answer

Bail, was right first time, c is not the speed of light its the speed of the electron.

anonymous · Answer

Bail on the bail, c IS just a constant which comes from Virial theorem.